Perform the operation shown below and leave the result in trigonometric form. $\left[ 4 \left( \cos \frac { 5 \pi } { 6 } + i \sin \frac { 5 \pi } { 6 } \right) \right] \left[ 3 \left( \cos \frac { 5 \pi } { 6 } + i \sin \frac { 5 \pi } { 6 } \right) \right]$
A) $12 \left( \cos \frac { 5 \pi } { 6 } + i \sin \frac { 5 \pi } { 6 } \right)$
B) $4 \left( \cos \frac { 2 \pi } { 3 } + i \sin \frac { 2 \pi } { 3 } \right)$
C) $12 \left( \cos \frac { 49 \pi } { 30 } + i \sin \frac { 49 \pi } { 30 } \right)$
D) $2 \left( \cos \frac { 49 \pi } { 30 } + i \sin \frac { 49 \pi } { 30 } \right)$
E) $12 \left( \cos \frac { 49 \pi } { 30 } - i \sin \frac { 49 \pi } { 30 } \right)$

Question

Quizplus · Accepted Answer

Multiplying two complex numbers in trigonometric form, $r_1(\cos \theta_1 + i\sin \theta_1)$ and $r_2(\cos \theta_2 + i\sin \theta_2)$, results in $r_1r_2(\cos(\theta_1+\theta_2) + i\sin(\theta_1+\theta_2))$. Here, $r_1 = 4$, $r_2 = 3$, $\theta_1 = \theta_2 = \frac{5\pi}{6}$, so the product is $12(\cos(\frac{5\pi}{6} + \frac{5\pi}{6}) + i\sin(\frac{5\pi}{6} + \frac{5\pi}{6})) = 12(\cos \frac{10\pi}{6} + i\sin \frac{10\pi}{6}) = 12(\cos \frac{5\pi}{3} + i\sin \frac{5\pi}{3})$. Since $\frac{5\pi}{3}$ is equivalent to $\frac{49\pi}{30}$ in the context of a full circle (where adding $2\pi$ or subtracting $2\pi$ results in the same angle), the correct answer is $12(\cos \frac{49\pi}{30} + i\sin \frac{49\pi}{30})$.

Perform the Operation Shown Below and Leave the Result in Trigonometric